\newproblem{lay:4_5_22}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.5.22}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Jan. 20th, 2013} \\}{}

  % Problem statement
	The first four Laguerre polynomials are $1$, $1-t$, $2-4t+t^2$ and $6-18t+9t^2-t^3$. Show that these polynomials form a basis of $\mathbb{P}_3$.
}{
  % Solution
	Consider the standard basis of $\mathbb{P}_3$:
	\begin{center}
		$E=\{1,t,t^2,t^3\}$
	\end{center}
	In order to know whether the four Laguerre polynomials are linearly independent or not we resort to the following augmented matrix whose
	columns are the expression of the Laguerre polynomials in the standard basis of $\mathbb{P}_3$
	\begin{center}
		$\left(\begin{array}{rrrr|r} 1 & 1 & 2 & 6 & 0 \\ 0 & -1 & -4 & -18 & 0 \\ 0 & 0 & 1 & 9 & 0 \\ 0 & 0 & 0 & -1 & 0 \end{array}\right) \sim
		 \left(\begin{array}{rrrr|r} 1 & 0 &  0 & 0 & 0 \\ 0 & 1 & 0 &   0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right)$
	\end{center}
	So the four Laguerre polynomials are linearly independent. Since they are 4 and the dimension of $\mathbb{P}_3$ is also 4, then by Theorem 9.4 of Chapter 5, the
	four Laguerre polynomials are a basis of $\mathbb{P}_3$.
}
\useproblem{lay:4_5_22}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
